LeetCode 1385. Find the Distance Value Between Two Arrays两个数组间的距离值【Easy】【Python】【暴力】
Problem
Given two integer arrays arr1
and arr2
, and the integer d
, return the distance value between the two arrays.
The distance value is defined as the number of elements arr1[i]
such that there is not any element arr2[j]
where |arr1[i]-arr2[j]| <= d
.
Example 1:
Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
Output: 2
Explanation:
For arr1[0]=4 we have:
|4-10|=6 > d=2
|4-9|=5 > d=2
|4-1|=3 > d=2
|4-8|=4 > d=2
For arr1[1]=5 we have:
|5-10|=5 > d=2
|5-9|=4 > d=2
|5-1|=4 > d=2
|5-8|=3 > d=2
For arr1[2]=8 we have:
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2
Example 2:
Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
Output: 2
Example 3:
Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
Output: 1
Constraints:
1 <= arr1.length, arr2.length <= 500
-10^3 <= arr1[i], arr2[j] <= 10^3
0 <= d <= 100
问题
给你两个整数数组 arr1 , arr2 和一个整数 d ,请你返回两个数组之间的 距离值 。
「距离值」 定义为符合此描述的元素数目:对于元素 arr1[i] ,不存在任何元素 arr2[j] 满足 |arr1[i]-arr2[j]| <= d 。
示例 1:
输入:arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
输出:2
解释:
对于 arr1[0]=4 我们有:
|4-10|=6 > d=2
|4-9|=5 > d=2
|4-1|=3 > d=2
|4-8|=4 > d=2
对于 arr1[1]=5 我们有:
|5-10|=5 > d=2
|5-9|=4 > d=2
|5-1|=4 > d=2
|5-8|=3 > d=2
对于 arr1[2]=8 我们有:
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2
示例 2:
输入:arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
输出:2
示例 3:
输入:arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
输出:1
提示:
1 <= arr1.length, arr2.length <= 500
-10^3 <= arr1[i], arr2[j] <= 10^3
0 <= d <= 100
思路
暴力
时间复杂度: O(n^2)
空间复杂度: O(1)
Python3代码
from typing import List
class Solution:
def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
# solution one: 暴力
res = 0
for x in arr1:
cnt = 0
for y in arr2:
if abs(x-y) <= d:
break
else:
cnt += 1
if cnt == len(arr2):
res += 1
return res
# solution two: 一行代码
return sum(all(abs(a1 - a2) > d for a2 in arr2) for a1 in arr1)
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